The industrial process for making the ammonia used in fertilizer, explosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure:

$N_{2}(g) + 3 H_{2}(g) → 2 NH_{3}(g)$

If 18.20 kg of $NH_{3}$ is produced by a reaction mixture that initially contains 6.00 kg of $H_{2}$ and an excess of $N_{2}$ , what is the percent yield of the reaction?

Question

The industrial process for making the ammonia used in fertilizer, explosives, and many other products is based on the reaction between nitrogen and hydrogen at high temperature and pressure:

[latex]N_{2}(g) + 3 H_{2}(g) → 2 NH_{3}(g)[/latex]

If 18.20 kg of [latex]NH_{3}[/latex] is produced by a reaction mixture that initially contains 6.00 kg of [latex]H_{2}[/latex] and an excess of [latex]N_{2}[/latex], what is the percent yield of the reaction?

Accepted Answer

Collect and Organize We know that the actual yield of [latex]NH_{3}[/latex] is 18.20 kg. We also know that the reaction mixture initially contained 6.00 kg of [latex]H_{2}[/latex] and that [latex]H_{2}[/latex] must be the limiting reactant because the problem specifies the presence of excess [latex]N_{2}[/latex]. We need to find the percent yield.

Analyze We need to use the mass of [latex]H_{2}[/latex] to calculate how much [latex]NH_{3}[/latex] could have been produced— the theoretical yield. We can then use the theoretical yield and the actual yield to calculate percent yield. We need the molar masses of [latex]H_{2}[/latex] and [latex]NH_{3}[/latex] and the stoichiometry of the reaction, which tells us that 2 moles of [latex]NH_{3}[/latex] are produced for every 3 moles of [latex]H_{2}[/latex] consumed.

Solve The molar masses we need are:

[latex]\mathscr{M}_{H_{2}} = 2(1.008 g/mol) =2.016 g/mol[/latex]

[latex]\mathscr{M}_{NH_{3}} =14.01 g/mol + 3(1.008 g/mol) = 17.03 g/mol[/latex]

We calculate the theoretical yield of [latex]NH_{3}[/latex]:

[latex]6.00 \sout{kg H_{2}} imes \frac{10^{3} \sout{g H_{2}}}{1 \sout{kg H_{2}}} imes \frac{1 \sout{mol H_{2}}}{2.016 \sout{g H_{2}}} imes \frac{2 \sout{mol NH_{3}}}{3 \sout{mol H_{2}}} imes \frac{17.03 \sout{g NH_{3}}}{1 \sout{mol NH_{3}}} imes \frac{1 kg NH_{3}}{10^{3} \sout{g NH_{3}}} =33.8 kg NH_{3}[/latex]

Then we divide the actual yield by this theoretical yield to determine the percent yield:

[latex]\frac{18.20 \sout{kg NH_{3}}}{33.8 \sout{kg NH_{3}}} imes 100 \%=53.8\%[/latex]

Think About It A yield of about 54% may seem low, but it may be the best that can be achieved for a particular process. A great deal of chemical research goes into trying to improve the percent yield of industrial chemical reactions.

Q. 3.22

Chapter 3

Q. 3.22

This Question has been Answered.

Already have an account?

Login

Related Answered Questions

One way of producing hydrogen for use in hydrogen powered vehicles is the water–gas shift reaction: H2O(g) + CO(g) → H2(g) + CO2(g) At 200°C, the reaction produces a 96% yield. How many grams of H2O and CO are needed to generate 176 g of H2 under these conditions? ...

Eugenol is an ingredient in several spices, including bay leaves and cloves (Figure 3.36). Eugenol contains carbon, hydrogen, and oxygen. Combustion of 21.80 mg of eugenol yields 58.5 mg of CO2 and 14.4 mg of H2O. What is the empirical formula of eugenol? If the mass spectrum of eugenol shows a ...

Nitrous oxide, the laughing gas used in dentists’ offices, is 63.65% N. What is the empirical formula of nitrous oxide? ...

The compound C2H2F4 is a propellant in the inhalers used by asthma sufferers (Figure 3.23). What is the percent composition of C2H2F4? ...